12. \( \int_{1}^{e} x^{4} \ln x d x \)

To solve the integral \( \int_{1}^{e} x^{4} \ln x \, dx \), we can use integration by parts. Let's set \( u = \ln x \) which gives us \( du = \frac{1}{x} dx \) or \( dx = x \, du \). We can also let \( dv = x^4 \, dx \), leading to \( v = \frac{x^5}{5} \). Using integration by parts: \[ \int u \, dv = uv - \int v \, du \] we have: \[ \int x^4 \ln x \, dx = \ln x \cdot \frac{x^5}{5} - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx \] \[ = \ln x \cdot \frac{x^5}{5} - \frac{1}{5} \int x^4 \, dx \] Calculating \( \int x^4 \, dx \), gives us \( \frac{x^5}{5} \). Therefore, we can substitute this back into our equation: \[ = \ln x \cdot \frac{x^5}{5} - \frac{1}{5} \cdot \frac{x^5}{5} = \ln x \cdot \frac{x^5}{5} - \frac{x^5}{25} \] Now, we need to evaluate this expression from \( 1 \) to \( e \). \[ \left[ \ln x \cdot \frac{x^5}{5} - \frac{x^5}{25} \right]_{1}^{e} \] Calculating at the bounds: At \( x = e \): \[ \ln(e) \cdot \frac{e^5}{5} - \frac{e^5}{25} = \frac{e^5}{5} - \frac{e^5}{25} = \frac{5e^5}{25} - \frac{e^5}{25} = \frac{4e^5}{25} \] At \( x = 1 \): \[ \ln(1) \cdot \frac{1^5}{5} - \frac{1^5}{25} = 0 - \frac{1}{25} = -\frac{1}{25} \] Now we combine these results: \[ \frac{4e^5}{25} - \left( -\frac{1}{25} \right) = \frac{4e^5}{25} + \frac{1}{25} = \frac{4e^5 + 1}{25} \] Thus, the final answer is: \[ \int_{1}^{e} x^{4} \ln x \, dx = \frac{4e^5 + 1}{25} \]